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bezimeni [28]
3 years ago
12

You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas

ing the area of the plates. D. Increasing the area of the plates. E. Both A and D F. Both B and C
Physics
1 answer:
eimsori [14]3 years ago
4 0

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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Starting at 9a.m., you ride your hover board for 3hrs at an average speed of 6 mph. Out of breath, you stop for tea from noon un
Elena-2011 [213]
3 times 6= 18. The average speed is 19 mph.

hope this helps!
6 0
2 years ago
Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h If doubled, Malcolm's maximum speed would be
AfilCa [17]

Answer

Given,

Average speed of Malcolm and Ravi = 260 km/h

Let speed of the Malcolm be X and speed of the Ravi Y.

From the given statement

\dfrac{X+Y}{2}=260

X + Y = 520 ....(i)

2X - Y = 80  ....(ii)

Adding both the equations

3 X = 600

 X = 200 km/h

Putting value in equation (i)

Y = 520 - 200

Y = 320 Km/h

Speed of Malcolm = 200 Km/h

Speed of Ravi = 320 Km/h

8 0
3 years ago
Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a
andrey2020 [161]

Explanation:

It is given that,

Force, F=(-8\ N)i+(6\ N)j

Position vector, r=(3i+4j)\ m

(a) The torque on the particle about the origin is given by :

\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m

(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
A container is filled to a depth of 19.0 cm with water. On top of the water floats a 31.0-cm-thick layer of oil with specific gr
IceJOKER [234]

Answer:

Absolute pressure of the oil will be 102822.8 Pa  

Explanation:

We have given height h = 31 cm = 0.31 m

Acceleration due to gravity g=9.8m/sec^2

Specific gravity of oil = 0.600

So density of oil \rho =0.6\times 1000=600kg/m^3

We know that absolute pressure is given by P=P_0+\rho gh, here P_0=1.01\times 10^5Pa

So absolute pressure will be equal to P=1.01\times 10^5+600\times 9.8\times 0.31=102822.8Pa

So absolute pressure of the oil will be 102822.8 Pa

6 0
3 years ago
43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a
olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N

The frictional force that acts on the sliding bear is 378.47166 N

5 0
3 years ago
Read 2 more answers
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